Let .
If is prime, we've done.
Else, let be a prime divisor of .
Then .
Let .
Let . Then .
But .
So .

If I am reading this right, then, the only case left to show is to prove that no group can exist where the order of all of its elements is 1. (I admit this is practically trivial, but it wasn't addressed in the above proof.)

If I am reading this right, then, the only case left to show is to prove that no group can exist where the order of all of its elements is 1. (I admit this is practically trivial, but it wasn't addressed in the above proof.)

-Dan

The only element of order 1 is the identity. He said "with more than 1 element (identity)".